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Well, I attended this years ARML competition, and I found the questions to be challenging, and the power round to be a lot of fun.

For those who don't know: arml.com

Basically, I'm posting this year's power round. I don't know the answers yet either, I'd like to see who can come up with something though... =D

ARML POWER QUESTION 2008 - Piles of Tiles

------------------------------------------------------------- (Rules of the power round at ARML(Applies here too, but not all of them(Specifically, only the bolded rules apply)))

The power question is worth 20 points. Each problem is worth 4 points. To receive full credit the presentation must be legible(not a problem here), orderly, clear, and concise. Even if not proved, earlier numbered items may be used in solutions to later numbered items but not vice versa. The pages submitted for credit should be NUMBERED IN CONSECUTIVE ORDER AT THE TOP OF EACH PAGE in what your team considers to be proper sequential order. PLEASE WRITE ON ONLY ONE SIDE OF THE ANSWER PAPERS.

Put the TEAM NUMBER (not the Team name) on the cover sheet used as the first page of the papers submitted. Do not identify the Team in any other way.

------------------------------------------------------------- (Some background information for the questions)

Given a rectangle whose dimensions are a x b, where a and b are integers, a tiling by rectangular tiles of dimensions u1 x v1, u2 x v2, etc., where all dimensions are integers, is an arrangement of tiles that fill the rectangle with no gaps and no overlaps. The edges of the tiles are parallel to the sides of the rectangle. For example, some of the ways to tiles a 4x4 rectangle with 2 x 2 and 1 x 4 tiles would be to use four of either tile shown by the left two diagrams or two of each as shown by the right two diagrams:


---- --- --- ---- ---- --- --- ---- ---- --- --- ---- ---- --- --- ----
| | | | | | | | |
---- --- --- ---- ---- --- --- ---- ---- --- --- ---- ---- ----
| | | | | | | | | | | |
---- --- --- ---- ---- --- --- ---- ---- --- --- ---- ---- --- ----
| | | | | | | | | | | | | | |
---- --- --- ---- ---- --- --- ----
| | | | | | | | | | | | | |
---- --- --- ---- ---- --- --- ---- ---- --- --- ---- ---- --- --- ---- ---- --- ---- ---- ----[/codebox]

((Sorry, I'm too lazy to upload an image. For each 'rectangle' lines indicate the actual lines that are drawn in and spaces are used to indicated the grid, unless filled in by a line.))

A [i]u[/i] x [i]v[/i] tile has a height of [i]u[/i] and a width of [i]v[/i]. The diagram shows a 2 x 3 tile on the left and a 3 x 2 tile on the right. Tiles cannot be rotated into different orientations and two [i]u x v[/i] are indistinguishable. As example of different tiling, there are two ways to tile a 1 x 3 rectangle with a 1 x 1 tile and a 1 x 2 tile---the 1x1 tile can go to the left or to the right of the 1 x 2 tile. Let S([i]n[/i]) and |S([i]n[/i])| represent the set of tilings of a 1 x [i]n[/i] rectangle and the number of such tilings respectively.

------------------------------------------------------------- (Now the actual questions.)

1. a) Compute the number of ways a 1 x 4 rectangle can be tile with 1 x 1 and 1 x 2 tiles.

b) Give two different tilings of a 12 x 15 rectangle with combinations of 2 x 4 and 3 x 3 tiles.

c) Prove that it is impossible to tile an 80 x 80 rectangle with tiles of sizes 3 x 11 and 4 x 15.

2. a) For [i]n[/i] that is an element of {1, 2, ..., 10}, compute the number of ways to tile a 1 x [i]n[/i] rectangle with tiles of sizes 1 x 1 and 1 x 2 tiles. Give your answers in a table.

b) Determine, with proof, a recursive formula to compute the number of ways to tile a 1 x [i]n[/i] rectangle with 1 x 1 and 1 x 2 tiles. use the formula to compute the number of ways to tile a 1 x 17 rectangle.

3. a) In a table, list all values of [i]n[/i] less than or equal to 40 such that a 1 x [i]n[/i] rectangle can be tiled with 1 x 10 and 1 by 12 tiles. For each [i]n[/i], list a combination of 10's and 12's that work.

b) Compute the number of values of [i]n[/i], where [i]n[/i] is greater than or equal to 1000 and less than or equal to 2008, for which a 1 x [i]n[/i] rectangle can be tiles with 1 x 10 and 1 x 12 tiles.

c) Determine, with justification, a recursive formula for the number of ways to tile a 1 x [i]n[/i] rectangle with 1 x 2 and 1 x 3 tiles. Then compute the number of ways to tile a 1 x 17 rectangle.

4. For this question only, suppose that 1 x 1 tiles are gray and 1 x 2 tiles are colored in two ways: the right half white and the left half black or vice-versa. Thus there are two distinct types of 1 x 2 tiles.

a) Using there colored tiles, determine the number of distinguishable tilings for a 1 x 17 rectangle. Explain the method used (we always do that, right? =P).

b) Using these colored tiles, give an [u]explicit, closed-form[/u] formula in terms of [i]n[/i] for the number of distinguishable tilings of a 1 x [i]n[/i] rectangle. This is a non recursive formula. Justify your formula.

5. Find, with justification, the number of different ways to tile an 8 x 8 rectangle using 2 x 2 and 1 x 3 tiles. It is not necessary to use both types in a tiling.

------------------------------------------------------------- (This exists on the paper. Seriously.)

For the rest of the problems, you may assume the following [i]Spanning Theorem[/i]: A 1 x [i]n[/i] rectangle can be tiles with tiles of sizes 1 x [i]u[/i] and 1 x [i]v[/i] if and only if there exist nonnegative integers [i]t[/i][sub]1[/sub] and [i]t[/i][sub]2[/sub] such that [i]ut[/i][sub]1[/sub] + [i]vt[/i][sub]2[/sub] = [i]n[/i]. Let #[i]u, v[/i]# denote {[i]ut[/i][sub]1[/sub] + [i]vt[/i][sub]2[/sub] : [i]t1,t2[/i] nonnegative integers}. We express this theorem compactly as follows:

A 1 x [i]n[/i] rectangle can be tiled with sizes 1 x [i]u[/i] and 1 x [i]v[/i] if and only if [i]n[/i] is an element of #[i]u, v[/i]#.

We now consider a generalization of the Spanning Theorem:

6. a) Prove that if an [i]m[/i] x [i]n[/i] rectangle can be tiled with [i]u[/i][sub]1[/sub] x [i]v[/i][sub]1[/sub] and [i]u[/i][sub]2[/sub] x [i]v[/i][sub]2[/sub] tiles, then [i]m[/i] is an element of #[i]u[/i][sub]1[/sub], [i]u[/i][sub]2[/sub]# and n is an element of #[i]v[/i][sub]1[/sub],[i] v[/i][sub]2[/sub]#. (*)

b) Give an example to show that the condition [i]m[/i] is an element of #[i]u[/i][sub]1[/sub], [i]u[/i][sub]2[/sub]# and n is an element of #[i]v[/i][sub]1[/sub],[i] v[/i][sub]2[/sub]# is not sufficient to guarantee the existence of a tiling of an [i]m[/i] x [i]n[/i] rectangle with [i]u[/i][sub]1[/sub] x [i]v[/i][sub]1[/sub] and [i]u[/i][sub]2[/sub] x [i]v[/i][sub]2[/sub] tiles.

----------------------------------------------------------- (I don't know why this exists on the paper either.)

Because the generalized spanning condition (*) does not guarantee the existence of a tiling of a given rectangle using two given tiles, most of the rest of this power question is devoted to understanding when such a tiling is possible.

Below are two tilings of a 6 x 7 rectangle with combinations of 2 x 2 and 1 x 3 tiles:

[codebox] ---- --- --- --- --- --- ---- ---- --- --- --- --- --- ----
| | | | | | | |
--- --- ---- --- --- ---
| | | | | | | |
---- --- --- --- --- --- ---- ---- --- --- --- --- --- ----
| | | | | | | |
--- --- ---- --- --- ---
| | | | | | | |
---- --- --- --- --- --- ---- ---- --- --- --- --- --- ----
| | | | | | | |
--- --- ---- ---- --- ---
| | | | | | | |
---- --- --- --- --- --- ---- ---- --- --- --- --- --- ----

Notice that in Tiling 1, the 6 x 7 rectangle has been effectively subdivided into two smaller rectangles of dimensions 6 x 4 and 6 x 3, each of which is tiled with copies of a single tile. To divide Tiling 2 into sub rectangles tiled with copies of a single tile would required seven sub rectangles: one 4 x 2, two 2 x 2's, three 2 x 3's and one 2 x 4. Call the complexity of a particular tiling the smallest number c, c greater than equal to 1, of sub rectangles, each tiled with copies of a single tile, into which a tiled rectangle can be divided. So Tiling 1 is of complexity 2 and Tiling 2 is of complexity 7.

In what follows, we will assume that no tile is a scaled copy of any other tile, that is, if one tile is u1 x v1 and the other tile is u2 x v2, we assume...

SORRY, I really need to get off now. I'll finish this tomorrow, and now, you can try the beginning. ;)

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All I have time for now is #1. So here goes:

5 ways

post-6822-1212413279_thumbjpg

Basically, the problem is that you can't get to 80 by using combinations of 11 and 15. All multiples of 15 end in either 0 or 5, so you would need a multiple of 11 to also end in 0 or 5 to make the two add to 80. The only way to do it then would be to use either 0 or 5 3x11 tiles. If you use 0 3x11 tiles, it won't work, because 15 * 5 = 75 and 15 * 6 = 90. If you use 5 3x11 tiles, you need some multiple of 15 to equal 35 (80 - 55). 15 * 2 = 30 which is too small and 15 * 3 = 45 which is too big. Therefore it cannot be done.

Here are two possibilities.

post-6822-1212413341_thumbjpg

I kind of feel like I'm cheating with the second one, but the rules didn't say I couldn't do it this way...

post-6822-1212413351_thumbjpg

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SORRY, I really need to get off now.

That's what she said. :)

Ah, good ol' ARML...I went back in 2003 and 2004 for Team North Carolina.

I think I've got #1, 2, and 3...

Pretty sure it's 5:

1a.png

1b.png

Since we can't rotate the tiles, there would need to be some combination of 11's and 15's that would add up to 80. That is, we need non-negative integers m and n such that 11m + 15n = 80. By inspection, this is not possible, so we cannot tile the 80 x 80 square with 3 x 11 and 4 x 15 tiles.

By inspection, |S(1)| = 1. Also, |S(0)| = 1.

For any n > 1, consider the rightmost tile in a tiling of a 1 x n rectangle. It must either be a 1x1 or a 1x2, which are two disjoint cases. In the 1x1 case, the number of ways to tile the rest of the rectangle is |S(n-1)| and in the 1x2 case, the number of ways to tile the rest of the rectangle is |S(n-2)|. Since these are the only possible cases, and the cases are disjoint, the number of ways to tile a 1xn rectangle with 1x1 and 1x2 tiles is |S(n)| = |S(n-1)| + |S(n-2)|. From this formula, we can find that |S(17)| = 2584.

N |S(n)|

1 1

2 2

3 3

4 5

5 8

6 13

7 21

8 34

9 55

10 89

So we can't use the spanning theorem? It seems like common knowledge...or at least the proof is pretty simple...

Suppose we are trying to tile a 1xn rectangle with 1xt1 and 1xt2 tiles. Suppose such a tiling is possible and the number of 1xt1 tiles we use is u, and the number of t2 tiles we use is v. Then, clearly u*t1 + v*t2 = n. Therefore, if a tiling is possible, then there exists non-negative u and v such that u*t1 + v*t2 = n.

Suppose now that such a tiling were not possible and that there existed non-negative integers u and v such that u*t1 + v*t2 = n. We can arrange u 1xt1 tiles followed by v 1xt2 tiles. This would be a tiling of the 1xn rectangle, which is a contradiction. Thus, if a such a tiling were not possible, there cannot exist non-negative integers u and v such that u*t1 + v*t2 = n.

Therefore, a tiling of a 1xn rectangle with 1xt1 and 1xt2 tiles is possible if and only if there exist non-negative integers u and v such that u*t1 + v*t2 = n.

Now we can use the spanning theorem.

For any integer n = 1, 2, ..., 40, in order for a tiling of a 1 x n rectangle using 1 x 10 and 1 x 12 tiles, we need non-negative integers a and b such that 10a + 12b = n. Non-negative n less than or equal to 40 for which this is possible are 0, 10, 20, 30, 40, 12, 24, 36, 22, 34, 32.

I claim that for all n = 1000, 1002, 1004, 1006, ... (all even integers greater than or equal to 1000), a 1xn rectangle can be tiled with 1x10 and 1x12 tiles. Clearly, for no odd n is this true. For all n that are elements of {1000, 1002, 1004, ..., 2008}, n = 0, 2, 4, 6, or 8 (mod 10). For these respective cases, we can use 0, 1, 2, 3, or 4 1x12 tiles, and fill in the rest of the 1xn rectangle with 1x10 tiles. Thus, the answer is the number of even integers between 1000 and 2008 inclusive, which is 505.

Similar to 2b, the tiling either ends with a 1x2 or 1x3 tile, so if T(N) is the number of ways to tile a 1xN rectangle with 1x2 and 1x3 rectangles, then T(0) = 1, T(1) = 0, T(2) = 1, T(3) = 1, T(4) = 1, and T(N) = T(N-2) + T(N-3) for integers N>4.

T(17) is therefore 49.

How did I do?

Don't feel like trying to figure out 4 or 5 right now.

Edited by Chuck
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Oh ehm gee. Darn, I don't have a lot of time today either.

For 1a, b, and c, you guys both have it right (though HoustonHokie, you mixed up b with c. =D)

I'm not even sure of the answers myself though, because ARML was only about two days ago, and the answers haven't come out yet. =O

And our team got answers for about the first five, just I don't know for sure, and i didn't really get to see anything besides number one and five. =P

As soon as I get the answers (this saturday?) I will confirm answers... if you guys can solve these problems that is. =P

The rest of the questions will come up... ASAP. Because now I don't know when I'll get to type the rest up.

((Chuck, nice job. =O Even though I don't know if the answers are right, you got pretty far on your own.(I wonder how long it took?)))

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((Chuck, nice job. =O Even though I don't know if the answers are right, you got pretty far on your own.(I wonder how long it took?)))

To solve them on paper...only about 20 minutes, but it took me a bit longer to type them out. Not entirely sure on all of them, but the recursive formula ones...I brute forced the first few and figured my reasoning was sound.

What team were you on?

Edited by Chuck
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Oh ehm gee. Darn, I don't have a lot of time today either.

For 1a, b, and c, you guys both have it right (though HoustonHokie, you mixed up b with c. =D)

I'm not even sure of the answers myself though, because ARML was only about two days ago, and the answers haven't come out yet. =O

And our team got answers for about the first five, just I don't know for sure, and i didn't really get to see anything besides number one and five. =P

As soon as I get the answers (this saturday?) I will confirm answers... if you guys can solve these problems that is. =P

The rest of the questions will come up... ASAP. Because now I don't know when I'll get to type the rest up.

((Chuck, nice job. =O Even though I don't know if the answers are right, you got pretty far on your own.(I wonder how long it took?)))

Oh, and out of curiosity, do you happen to have the individual round?

-Chuck

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Oh, and out of curiosity, do you happen to have the individual round?

-Chuck

-shivers-

I hated that individual round.

I'm sure I lost the papers somewhere while we were traveling, but I do remember what most of the questions were.

Team? AAST Frosh A. Horrible team we are. But our school always brings the most people. XD So we're the loudest if we win anything.

>_> Again, sorry, I had a concert today, so no time again.

Huh... I wonder why no one else attempts the power round?

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-shivers-

I hated that individual round.

I'm sure I lost the papers somewhere while we were traveling, but I do remember what most of the questions were.

Meh...I went to ARML twice and got a total of 4 questions right in two years, so I hated it too. Although our team did win the super relay one year. That was pretty epic.

Team? AAST Frosh A. Horrible team we are. But our school always brings the most people. XD So we're the loudest if we win anything.

>_> Again, sorry, I had a concert today, so no time again.

What is AAST?

Huh... I wonder why no one else attempts the power round?

Because it takes more than 3 minutes to figure it out and most of the ppl on this forum are not on summer vacation? And even if they are, they'd rather not do math?

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... I don't think the forum likes me. T_T It's not letting me edit my posts.

Anyways, more of the power round!

...we assume u1/u2 is not equal to v1/v2.

7. Under what conditions on u1, v1, u2, v2, m, and n does an m x n rectangle have a tiling of complexity 1 when u1 x v1 and u2 x v2 tiles are used? Justify your answer.

8. Under what conditions on u1, v1, u2, v2, m, and n does an m x n rectangle have a tiling of complexity at most 2 when u1 x v1 and u2 x v2 tiles are used? Justify your answer.

Let R be an m x n rectangle, where m and n are not necessarily integers, that has been colored in a checkerboard pattern with squares of size 1/2 x 1/2, as in the diagram below, with m similar to 3.2 and n similar to 4.8.

((Um... can't draw it... maybe I'll draw it later.))

You may assume as a lemma that if a u x v tile is placed on R with edges parallel to the sides of the rectangle, and at least of of u and v is an integer, then the tile covers equal areas of black and white.

9. Prove that if R is tiled with two different kinds of tiles, where each kind of tile has at least one integral dimension, then at least one of m and n is an integer.

10. If u1, v1, u2, v2, m and n are all positive integers, and if an m x n rectangle R can be tiled with tiles of dimensions u1 x v1 and u2 x v2, prove that it has tiling with those tiles of complexity at most 2.

And that's the end. I might decide to post up individual's later.

Edited by PhoenixTears
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I think I've got 4, 7, and 8...I'm pretty close on 5, although I've spent more time than I'm supposed to spend on it. I didn't really feel like reading and digesting all of 6.

This one is pretty much the same method as #2. Suppose C(n) is the number of ways to tile a 1xn rectangle in the way described. By brute force, C(1) = 1 and C(2) = 3. If we want, we can define C(0) = 1, since there's technically one way to arrange no tiles at all (compare to combinatorics, when N choose 0 is 1 for any non-negative integer N). This also makes the rest of the problem work out nicely.

Again, every tiling must end with a 1x1 or a 1x2. Similar to #2, if a tiling ends with a 1x1, then the number of ways for this to happen is C(n-1) for n>1. However, if a tiling ends in a 1x2, the last tile can be arranged in two ways, so the number of ways for a tiling of a 1xn rectangle to end in a 1x2 is 2*C(n-2). Thus, a recurrence relation for C(n) is:

C(n) = C(n-1) + 2*C(n-2). Now, I used a method from a college discrete math class to figure out the closed form. I will assume a general closed form, solve for the constants, and then prove that it is correct. If you're confused about how I did this, search for "recurrence relation" on Wikipedia to see the general way to solve these for closed forms.

My guess:

C(n) = Arn, where A is not zero.

C(n) = C(n-1) + 2*C(n-2)

Substitute the general closed form:

Arn = Arn-1 + 2Arn-2

Divide out Arn-2:

r2 = r + 2

Solve the quadratic:

r = -1 or 2.

New guess:

C(n) = A*(-1)n + B*2n, where A and B are real numbers, and may or may not be zero. Plug in the initial conditions (C(1) = 1 and C(2) = 3) to solve for A and B:

C(1) = 1 = -A + 2B

C(2) = 3 = A + 4B

(A,B) = (1/3, 2/3)

Conjecture (and what will turn out to be the answer):

C(n) = (1/3)*(-1)n + (2/3)*2n

Now to prove by induction that this does in fact work for the recurrence relation:

Spoiler for If you really want to see the proof by induction...:

By inspection, we can verify this closed form for n = 0, 1, and 2. For C(3) = 5, it still works.

Suppose that for some positive integer k, that C(k) = (1/3)*(-1)k + (2/3)*2k and that C(k-1) = (1/3)*(-1)k-1 + (2/3)*2k-1.

For induction, we want to show that C(k+1) = (1/3)*(-1)k+1 + (2/3)*2k+1.

From the recurrence relation, we have:

C(k+1) = C(k) + 2*C(k-1)

C(k+1) = (1/3)*(-1)k + (2/3)*2k + 2*((1/3)*(-1)k-1 + (2/3)*2k-1)

Distributing and rearranging terms:

C(k+1) = (1/3)*(-1)k + (2/3)*(-1)k-1 + (2/3)*2k + (4/3)*2k-1

Consecutive powers of -1 cancel out (i.e. (-1)k + (-1)k+1 = 0). Also, get like powers of 2:

C(k+1) = (1/3)*(-1)k-1 + 2/3*2k + 2/3*2k

Multiply by the first term by 1, written in the sneaky form -1*-1, and combine the last two terms:

C(k+1) = (1/3)*(-1)k-1*-1*-1 + 2/3*2k+1

C(k+1) = (1/3)*(-1)k+1 + 2/3*2k+1

QED.

Plugging in 17 for n, I got the number of ways to tile a 1x17 rectangle with gray 1x1 and white/black 1x2 tiles is 87,381

Clearly for complexity 1, the entire rectangle must be made of copies of one tile. Our large rectangle is m x n, and our tiles are u1 x v1 and u2 x v2. Thus, for complexity 1, at least one of the following must be true:

  • u1 divides m AND v1 divides n
  • u2 divides m AND v2 divides n

I told you...I'll do it later...

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All right! If I'm right, these are the individual questions. It's not word for word, but I think this is fairly accurate enough. And these should be WAY easier than the power round.

1-1.

<------------------------------------------------->

						  1	/

							 /

						   /

						   \	2

							 \

						   3   \

							   /

							 /

						   /   4

<--------------------------------------------------->

((Note that the diagram is DEFINITELY not drawn to scale.))

If the m<1 < m<2 < m<4 < m<3, m<1 = 50, and m<3 < 90, find the greatest integer value for the m<3.

1-2. An integer is quadly when it has only four factors. Find the first number in a set of three consecutive quadly integers.

O_O And I forget the rest.

1 & 2 are pretty simple. Don't know about the rest of it.

EDIT: And... I don't have answers yet. T_T Sorry!

Edited by PhoenixTears
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All right! If I'm right, these are the individual questions. It's not word for word, but I think this is fairly accurate enough. And these should be WAY easier than the power round.

1-1.

<------------------------------------------------->

						  1	/

							 /

						   /

						   \	2

							 \

						   3   \

							   /

							 /

						   /   4

<--------------------------------------------------->

((Note that the diagram is DEFINITELY not drawn to scale.))

If the m<1 < m<2 < m<4 < m<3, m<1 = 50, and m<3 < 90, find the greatest integer value for the m<3.

1-2. An integer is quadly when it has only four factors. Find the first number in a set of three consecutive quadly integers.

O_O And I forget the rest.

1 & 2 are pretty simple. Don't know about the rest of it.

EDIT: And... I don't have answers yet. T_T Sorry!

I've spent more time than I should have.

1. I have no clue on the first one. Maybe I didn't interpret the picture correctly...is it just 4 random line segments spanning two parallel lines?

2. The three integers are 33, 34, 35. A quadly integer must be a product of two primes, or the 3rd power of a prime. I figured 53 would be too big and 33 doesn't work, so I just wrote down primes and took their products until I found 5*7 = 35, and 33 and 34 before it work. So the answer is 33.

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GRAR. Sorry. Number one I wasn't specific enough.

Label the two lines on the top and bottom lines l and m. They are parallel.

>_> Sorry.

I'll also see if I can get the rest of the stuff.

And yup, you're right for number 2 Chuck.

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